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23/05/2013 00:48:54

### Author Topic: Are some people born good at maths?  (Read 2956 times)

#### wolfekeeper

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• Reply #25 on: 04/02/2012 23:29:20
Um. Well delta x/delta t is the average speed over a time delta t. So for example delta t might be .1 second and the distance travelled might be 5 metres so the average speed would be 50 m/s.

If the delta is made very small, then you're averaging the speed over a very small time, then the average speed tends to the limit of the actual, instantaneous speed.

In other words for the exact speed, rather than just the close average, you want dx/dt which is the limit of delta-x/delta t.

So dx/dt is the speed. Geezer already calculated that (correctly!) for you.

So all you have to do is plug t=10 into Geezer's equation, and you have the exact speed at that moment.
« Last Edit: 04/02/2012 23:32:05 by wolfekeeper »

#### Geezer

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• Reply #26 on: 05/02/2012 06:05:17
Wow! My memory is not quite as bad as I thought it was.

I poked around a bit and came up with this http://www.youtube.com/watch?v=-duE_JQmjq8

This lady has gone to some trouble to explain an example in practical terms. (I confess I didn't sit through the entire lecture.)

#### CliffordK

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• Reply #27 on: 05/02/2012 11:44:38
Thanks,

Sometimes it is easiest to just do a numerical estimate which can be pretty easy to setup with a spreadsheet.

However, it is a good point that one can use derivatives and integrals to convert between distance, velocity, and acceleration.

A few days ago, there was a question...
Say one put a tunnel through the moon, and had magnetic acceleration, how fast could one get an object going.

I came up with an answer numerically with Opencalc, but I see that I should have done it mathematically if I wasn't a bit rusty.

So, now if I set my distance to be the diameter of the moon (3,475 km = 3,475,000 m) and solve for T, I get:

So, t=266.3 seconds.

Plugging that back into the velocity equation, one gets

v=10gt = 10*9.8*266.3 = 26097 m/s = 26 km/s

Now, let's see how I did.
Oops, I did the calcs for 100g (or 1000 m/s2)

But, it isn't a big deal...  just change my 5g to 50g above.

And, I get t=84.2 seconds (or 83.4 seconds if I used g=10m/s2)
V=100gt = 82.5 km/s

Anyway, so essentially my numerical estimate was the same.

But, perhaps not as elegant as using a few squiggles.

Anyway, you can use calculus to change from distance to velocity to acceleration, and visa versa.

Derivatives gives one slopes (or instantaneous rate of change) of a curve.
Integrals gives one the area under a curve.

Certainly it can be applied to many other things such as population models.

#### David Cooper

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• Reply #28 on: 06/02/2012 01:05:44
To clarify:-

Original equation for the position (distance travelled): x = 0.5 t^3 + 3 t^2

It now appears that differentiating that gives the equation for the speed: dx/dt = 1.5 t^2 + 6 t

(For anyone reading this who doesn't know how differentiation is done, you multiply the 0.5 by the 3 in "^3"  [which means cubed] and take 1 away from that 3, then multiply the later 3 by the 2 in "^2" [which means squared] and take 1 away from that 2 [to get "^1" which obviously doesn't need to be stated]. It's a startlingly simple process which I spent months doing in school without ever being allowed to know why.)

Um. Well delta x/delta t is the average speed over a time delta t. So for example delta t might be .1 second and the distance travelled might be 5 metres so the average speed would be 50 m/s.

As soon as these deltas come into it, I get confused because they seem to have no purpose - all it's actually saying is that distance divided by time = average speed.

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If the delta is made very small, then you're averaging the speed over a very small time, then the average speed tends to the limit of the actual, instantaneous speed.

Which it appears will always be zero because you're hacking it back to the speed at the start when it wan't moving.

Quote
In other words for the exact speed, rather than just the close average, you want dx/dt which is the limit of delta-x/delta t.

But here you apply the trick of differentiating the original equation to derive one to give the speed at any point in time - all you need to do then is put in a value for the time and solve it. That's dead easy, but you do have to know that this new equation will serve that purpose. Up until today, I didn't.

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So dx/dt is the speed. Geezer already calculated that (correctly!) for you.

So all you have to do is plug t=10 into Geezer's equation, and you have the exact speed at that moment.

I did that, and since you've confirmed that this is the right way to do things, I now know something that I should have been told at school. Thank you for filling in a massive hole in my education. I have finally seen a practical application for calculus, assuming that this is actually calculus (seems too simple).

#### David Cooper

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• Reply #29 on: 06/02/2012 01:13:41
I poked around a bit and came up with this http://www.youtube.com/watch?v=-duE_JQmjq8

This lady has gone to some trouble to explain an example in practical terms. (I confess I didn't sit through the entire lecture.)

I'll watch that next time I travel to somewhere with a fast enough internet connection. I'm stuck with something that keeps jumping between 3G and GPRS and struggles to download audio.

#### David Cooper

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• Reply #30 on: 06/02/2012 02:17:53

So, now if I set my distance to be the diameter of the moon (3,475 km = 3,475,000 m) and solve for T, I get:

So, t=266.3 seconds.

Plugging that back into the velocity equation, one gets

v=10gt = 10*9.8*266.3 = 26097 m/s = 26 km/s

I'm having trouble following that, partly because I can't tell whether letters are constants or units, and I don't know what the dt is doing.

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Anyway, you can use calculus to change from distance to velocity to acceleration, and visa versa.

Well, I can now go from the first to the second and back, but how do you deal with acceleration?

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Derivatives gives one slopes (or instantaneous rate of change) of a curve.
Integrals gives one the area under a curve.

Stated in that form with nothing to tie it to a practical application, that just goes in one ear and out the other. That is why so many people fail to learn calculus - it needs to be grounded in practical application first before it goes abstract. You need to have an understanding of what use it is to work out the area under a curve, and all of these basic things need to be constantly available somewhere for you to refer back to so that you can fix and refix it in your mind. Some teachers fling ideas out at a hundred miles an hour and never repeat them, and you can never tell what's important and what's just a passing comment. There should be a standard map of all this stuff which has the whole thing set out clearly and consicely so that it's easy to see what calculus actually is and what it's for. I'd like to create such a map, but I obviously don't know what to put in it yet.

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Certainly it can be applied to many other things such as population models.

If anyone reading this has applied calculus to a real problem that would further help illustrate its uses, please feel free to share it here - it strikes me that the process is not at all difficult, but that the real problem is understanding what the squiggles mean and knowing what use it is when you apply it to something. (That last sentence is not faulty - you have to read "use" as "yooss" rather than"yooz".)

Thanks for all the help so far - you have already unlocked a door for me, and I hope others are finding this helpful too.

#### Geezer

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• Reply #31 on: 06/02/2012 03:05:49

It's a startlingly simple process which I spent months doing in school without ever being allowed to know why.

Try this:

A population of bacteria squares every second. Let's call the population p, and the time t.

at time 1, p=1
at time 2, p=4
at time 3, p=9

(you can see where this is going)

If I have not mucked this up, p = t^2

You can draw a graph of what this might look like, but it should be obvious that the slope gets steeper with time. In fact, after a few seconds, the slope is almost vertical.

Differentiation allows you to determine just how steep a slope is at a particular point on the slope. If you plot this example on to a piece of graph paper, you will find that, for example, at time 3 seconds, the slope ratio is 9:1.5 which equals 6.

If you differentiate p = t^2 you get,

dp/dt = 2t

When t is 3, 2t = 6

It's obvious by inspection that the slope ratio is 9:1.5, but I haven't a clue why that is!
« Last Edit: 06/02/2012 03:09:52 by Geezer »

#### CliffordK

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• Reply #32 on: 06/02/2012 04:00:55
I'm having trouble following that, partly because I can't tell whether letters are constants or units, and I don't know what the dt is doing.

Ahh, yes,
I always hate it when letters are poorly defined.

Actually, in the equations I wrote, the only real variable is t = time.

g is the gravitational constant, 9.8 m/s2, where m is meters, and s is seconds.  km would be kilometers.
So, like pi, (π), g does not vary, although sometimes I'll round it to 10 m/s2.  Your gravity might vary a little based on your location, for example on the moon.  But, then it can be written as a fraction of Earth's gravity, g.

m = meters
s = seconds
km = kilometers

I am expressing Distance as a function of time Distance(t).
Likewise Velocity is a function of time Velocity(t).
And Acceleration is a function of time Acceleration(t).

d....  That signifies change.  in something.  You will also see delta (Δ), which means change in time, or the symbol (∂) for partial derivatives.

Why the dt?  I suppose it tells you that you are evaluating a function with respect to the change of time.  And, thus, it is required to be  a part of integration.

Hmmm
Relating Distance, Velocity, and Acceleration.

if  f(t) is a distance function with respect to time t.
Then Velocity(t) is a function of the rate of change of the distance at any time t.
I.E.  If you are driving from LA to New York.  If Velocity = 0, that means you are parked, and distance is not changing.
If the Velocity is > 0, then you are going forward.
If the Velocity is < 0, then you are going backward.

Anyway, the Velocity(t) is the rate of change of the distance function, or the derivative of the distance function d(f(x))

Acceleration tells you how fast you are changing the velocity function (how much you've pressed the gas pedal down).  So, it would be the slope of the velocity function at any given time.  Acceleration(t) = d(Velocity(t)) = d(d(f(t))

I suppose I forgot that when one takes derivatives, one looses one's constant factors.  So, taking integrals, one should add them back in when doing integrals.  I.E.  What is the initial distance or velocity, which in some cases can be set to zero.

Anyway, the mathematics is relevant for scientific discussions.  However, one can often understand a lot about a subject without needing anything very complex.

#### David Cooper

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• Reply #33 on: 06/02/2012 05:56:45
Try this:

A population of bacteria squares every second. Let's call the population p, and the time t...

Differentiation allows you to determine just how steep a slope is at a particular point on the slope.

I managed to follow that. The slope of the graph doesn't seem so immediately meaningful in human terms, unlike in a slope in a graph where the slope relates to the speed of travel (which is something that we can relate to well), but yes: it will be the speed of population growth at that point (if that really is how a population of something grows, though that doesn't seem likely - doubling every so many minutes/hours would I'm guessing be more realistic for bacteria, and with other kinds of creature it might follow the Fibonacci Sequence).

By the way, this online graphic calculator might come in handy if anyone needs to see what any of these equations looks like on a graph (it uses Java):-

http://www.coolmath.com/graphit/

#### David Cooper

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• Reply #34 on: 06/02/2012 07:10:26
Most of it's now clear, but there are a few bits that aren't:-

Velocity = acceleration x time, but here you're complifying it by putting in front and turning time into dt, then you remove the and the d disappears. I don't know what the rule is that allows you to do this, but it doesn't look all that simple as it gets more complicated the next time:-

Quote

Distance is the area under the graph if the graph is of the velocity, so I can see the reason for the in front of velocity this time. I'm trying to work out what the dt bit means here, but clearly it can't just be velocity on its own as it needs to be an equation of the velocity, so that appears to be the meaning of the dt. Where I then get lost is where the disappears and appears to take the dt with it instead of just the d as it did last time. This kind of manipulation of parts of equations is another area which my maths teacher failed to put across in anything remotely resembling a systematic way - she did it without explanations and made it seem like witchcraft.

if  f(t) is a distance function with respect to time t.

Straight away I feel ill - this is caused by a year of seeing "f(x)=" appearing all over the board for no obvious reason. It left me with a complex about it and anything that looks like it because it was never meaningful. This is why I'd like to ban all the squiggles in the early stages and do as much as possible using normal words.

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Then Velocity(t) is a function of the rate of change of the distance at any time t.
I.E.  If you are driving from LA to New York.  If Velocity = 0, that means you are parked, and distance is not changing.
If the Velocity is > 0, then you are going forward.
If the Velocity is < 0, then you are going backward.

Had to read that a couple of times, but then it clicked.

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Anyway, the Velocity(t) is the rate of change of the distance function, or the derivative of the distance function d(f(x))

If f(x) causes me problems (it makes my mind go blank), imagine what d(f(x)) does! I won't be able to handle these shorthands for things until I've understood it in normal language. It's hard enough when it's still in normal language without wanting to remove that and replace any of it with squiggles.

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Acceleration tells you how fast you are changing the velocity function (how much you've pressed the gas pedal down).  So, it would be the slope of the velocity function at any given time.  Acceleration(t) = d(Velocity(t)) = d(d(f(t))

And that's even worse. If you're taught this stuff the right way from the outset it may be easy, but if you've been taught it badly and have nightmares about it, there's a lot of damage that needs to be undone. I just can't handle the squiggles at this stage. The maths behind it all is clearly dead easy, but it's all being hidden behind a wall of squiggles which only get in the way of understanding. These squiggles are a shorthand for mathematicians to make things more complact - they are not appropriate as a way in.

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I suppose I forgot that when one takes derivatives, one looses one's constant factors.  So, taking integrals, one should add them back in when doing integrals.  I.E.  What is the initial distance or velocity, which in some cases can be set to zero.

I understand that without difficulty though.

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Anyway, the mathematics is relevant for scientific discussions.  However, one can often understand a lot about a subject without needing anything very complex.

Indeed one can, but there's no reason why anyone should be shut out of the maths - it is simplicity obscured by squiggles. What people really need to see to get an easy ride into calculus (or anything else of this kind) is clear rules laid out as to how to go about things, clear descriptions of what things are, and clear demonstrations of the procedures you have to go through to solve problems. If any of the tools aren't available to you or if any part of the language isn't clear to you, you're stuffed - you can appear to be a failure at maths, and yet that is not the case. The real problem is that you haven't been given a checklist of tools so that you can be sure you're working with the complete set, you haven't been given an adequate (or even an inadequate) manual explaining the squiggles which you can go back to consult at any time, and you're pushed into doing everything through squiggles before you're ready. It's no wonder that so many people get stuck here (or somewhere else in maths) and never get any further. The problem isn't with the learners, but with the teaching.

#### Geezer

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• Reply #35 on: 06/02/2012 08:24:16

Indeed one can, but there's no reason why anyone should be shut out of the maths - it is simplicity obscured by squiggles. What people really need to see to get an easy ride into calculus (or anything else of this kind) is clear rules laid out as to how to go about things, clear descriptions of what things are, and clear demonstrations of the procedures you have to go through to solve problems. If any of the tools aren't available to you or if any part of the language isn't clear to you, you're stuffed - you can appear to be a failure at maths, and yet that is not the case. The real problem is that you haven't been given a checklist of tools so that you can be sure you're working with the complete set, you haven't been given an adequate (or even an inadequate) manual explaining the squiggles which you can go back to consult at any time, and you're pushed into doing everything through squiggles before you're ready. It's no wonder that so many people get stuck here (or somewhere else in maths) and never get any further. The problem isn't with the learners, but with the teaching.

I think you are correct. Academia has no excuse for failing to make calculus as easy to understand as arithmetic.

#### RD

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• Reply #36 on: 06/02/2012 12:34:58
Q. Are some people born good at maths?

some people are born bad at maths ... http://en.wikipedia.org/wiki/Discalculia

#### Geezer

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• Reply #37 on: 07/02/2012 01:07:22
At the risk of hopelessly derailing the thread (not that that ever stopped me) I saw something recently about people seeing numbers in particular colo(u)rs, at least I think it was numbers.

#### David Cooper

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• Reply #38 on: 07/02/2012 21:41:22
Ah yes, coloured fridge magnets. There's also a phenomenon called synesthesia which you might find interesting - some people can hear musical notes as colours, and some perceive numbers, letters or words as having colours, or tastes. I don't know what implications it has for their ability to enjoy music or to read/calculate/etc.

#### wolfekeeper

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• Reply #39 on: 07/02/2012 21:51:26
Um. Well delta x/delta t is the average speed over a time delta t. So for example delta t might be .1 second and the distance travelled might be 5 metres so the average speed would be 50 m/s.

As soon as these deltas come into it, I get confused because they seem to have no purpose - all it's actually saying is that distance divided by time = average speed.
Nearly, it's the distance CHANGE and the time CHANGE.

distance change divided by time change = average speed over that time

i.e.

delta x / delta t = average speed

Quote
Quote
If the delta is made very small, then you're averaging the speed over a very small time, then the average speed tends to the limit of the actual, instantaneous speed.

Which it appears will always be zero because you're hacking it back to the speed at the start when it wan't moving.
Wrong!!!! And it's wrong in an important way.

These are DELTAS so it would be delta x (0) over delta t (0) = 0/0 = undefined

So you can't put it all the way back, because the maths blows up when you set the time delta to exactly zero, a zero sized triangle has no calculable gradient.

But by doing the LIMIT, you are effectively setting the time delta an infinitesimally small amount above zero, so it doesn't blow up, and you get a sensible answer. And that's how calculus works.
« Last Edit: 07/02/2012 21:58:54 by wolfekeeper »

#### RD

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• Reply #40 on: 07/02/2012 22:22:58
At the risk of hopelessly derailing the thread (not that that ever stopped me) I saw something recently about people seeing numbers in particular colo(u)rs, at least I think it was numbers.

#### David Cooper

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• Reply #41 on: 07/02/2012 22:26:22
Q. Are some people born good at maths?

some people are born bad at maths ... http://en.wikipedia.org/wiki/Discalculia

There's been a heated argument about dyslexia with some people claiming there's no such condition - I saw a documentary about this where they took dyslexic children and taught them how to read properly, thereby curing them completely and in a very short time. I wouldn't be surprised if the same argument is going on with dyscalculia, even though with both conditions there appears to be quite a weight of evidence to support them. The truth may be that most examples of people diagnosed with these conditions don't actually have them, but some may well do.

#### David Cooper

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• Reply #42 on: 07/02/2012 22:29:51

I wonder how they feel when looking at something like that if the colours are wrong for them.
« Last Edit: 07/02/2012 23:45:01 by David Cooper »

#### David Cooper

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• Reply #43 on: 07/02/2012 23:44:23
As soon as these deltas come into it, I get confused because they seem to have no purpose - all it's actually saying is that distance divided by time = average speed.
Nearly, it's the distance CHANGE and the time CHANGE.

distance change divided by time change = average speed over that time

i.e.

delta x / delta t = average speed

But it isn't a change in distance - it's a change in location, and the distance represents the amount of change in location. With "time" it's more ambiguous as you can see time both as a location and as a duration, the latter being a change of the former. So, it now looks to me as if it's saying distance/duration, or change in location / change in time-location. Either way, I still can't see why the delta is necessary because metres and seconds are primarily units of distance and duration rather than locations.

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Quote
Quote
If the delta is made very small, then you're averaging the speed over a very small time, then the average speed tends to the limit of the actual, instantaneous speed.

Which it appears will always be zero because you're hacking it back to the speed at the start when it wan't moving.
Wrong!!!! And it's wrong in an important way.

These are DELTAS so it would be delta x (0) over delta t (0) = 0/0 = undefined

So you can't put it all the way back, because the maths blows up when you set the time delta to exactly zero, a zero sized triangle has no calculable gradient.

But by doing the LIMIT, you are effectively setting the time delta an infinitesimally small amount above zero, so it doesn't blow up, and you get a sensible answer. And that's how calculus works.

My point was with that example simply that you can see from the graph that if you make it infinitely small the speed will be zero - it doesn't give you any way to calculate the speed at a more interesting point on the graph. Since then we've got beyond that and I've seen that you can derive a new graph from the original one and then pick any time you like, replcae t with it, solve the equation and out pops the precise speed for that point in time.

#### wolfekeeper

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• Reply #44 on: 08/02/2012 00:20:29
But it isn't a change in distance - it's a change in location, and the distance represents the amount of change in location.
For example at t=3, you might be at x=4, 4 is a distance from the origin
and at t=3.3 you might be at 4.2, 4.2 is the distance from the origin

But delta x is NOT 4 or 4.2, no, it's the change in distance, the distance delta (delta literally just means change) in the distance: 4.2 - 4 = 0.2 = delta x when you make a small step in time

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Either way, I still can't see why the delta is necessary because metres and seconds are primarily units of distance and duration rather than locations.
They're also units of change/delta of distance and duration.

#### David Cooper

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• Reply #45 on: 08/02/2012 21:10:48
But it isn't a change in distance - it's a change in location, and the distance represents the amount of change in location.
For example at t=3, you might be at x=4, 4 is a distance from the origin
and at t=3.3 you might be at 4.2, 4.2 is the distance from the origin

But delta x is NOT 4 or 4.2, no, it's the change in distance, the distance delta (delta literally just means change) in the distance: 4.2 - 4 = 0.2 = delta x when you make a small step in time.

Well, it's a very strange way of describing something - the 0.2 is itself a distance between two locations, so to call it a change in distance seems unnecessary unless you're thinking about actively changing the distance between two points in some way by moving one or both of them along the graph.

#### wolfekeeper

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• Reply #46 on: 08/02/2012 22:10:34
It's not just any distance, it's specifically a very small distance, as small as possible.

#### Geezer

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• Reply #47 on: 08/02/2012 23:16:21
It's not just any distance, it's specifically a very small distance, as small as possible.

Right! You really could care less what the actual distance is because you are only interested in the ratio of the distances in two axes.

Perhaps you could substitute something along the lines of "the ratio of the axes at any point" for the shorthand "dx/dy"

#### imatfaal

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• Reply #48 on: 09/02/2012 12:01:15
It's not just any distance, it's specifically a very small distance, as small as possible.

Right! You really could care less what the actual distance is because you are only interested in the ratio of the distances in two axes.

Perhaps you could substitute something along the lines of "the ratio of the axes at any point" for the shorthand "dx/dy"
Technically , dy/dx is not a ratio - it is the limit or just a notation.  Mad mathematicians would tell you that the ratio of two infinitessimals can be anything you choose (kind of like the ration of two infinities it is not well defined).  It is a technical point and not that important - I have seen thinking of dy/dx as a ratio described as a "useful crutch that could get you into difficulties later".  The reasons are abstruse and arcane - and I couldn't rehearse them off hand (or possibly at all), but I have followed some very heated exchanges in which  mathematicians lose their rag trying to explain to the math forum visitor why it should not technically be thought of as ratio

#### wolfekeeper

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• Reply #49 on: 09/02/2012 14:48:17
Technically , dy/dx is not a ratio - it is the limit or just a notation.
Yes, that's correct, although if you ignore that, it turns out that you largely get away with it.
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Mad mathematicians would tell you that the ratio of two infinitessimals can be anything you choose (kind of like the ration of two infinities it is not well defined).
No, AFAIK that's definitely wrong. The reason that calculus works is that the LIMIT of dy/dx as you take the deltas towards zero is (in most normal cases) completely well defined. In some case limits can take two values in different directions where the curve jumps, so the curve has to be smooth where you differentiate; trying to differentiate a fractal doesn't do anything very good!

It's the number you get when you divide two numbers that are actually zero which is undefined, not the limit.
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It is a technical point and not that important - I have seen thinking of dy/dx as a ratio described as a "useful crutch that could get you into difficulties later".  The reasons are abstruse and arcane - and I couldn't rehearse them off hand (or possibly at all), but I have followed some very heated exchanges in which  mathematicians lose their rag trying to explain to the math forum visitor why it should not technically be thought of as ratio
Yeah, probably, dy/dx is really a notational thing, and there's some spectacular ways you can get into trouble with not realising this. Cancelling the 'd's is the least of your problems if you don't realise it(!) Also d^2y/dx^2 is even more obviously different. You can at least multiply dy/dx by dx and it makes sense and gives you dy, whereas multiplying d^2y/dx^2 by stuff doesn't do anything very good at all so far as I recall.
« Last Edit: 09/02/2012 14:52:06 by wolfekeeper »

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